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umka21 [38]
2 years ago
11

17

Physics
1 answer:
satela [25.4K]2 years ago
8 0

Answer:

all the lamps has the same brightness

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Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
soldi70 [24.7K]
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
2 years ago
If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
steposvetlana [31]

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0
2 years ago
A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar
____ [38]

Answer:

60

so you take 120÷2 wires

4 0
2 years ago
Gravitational Potential Energy = mass x gravity x height
Blizzard [7]
M= gpe / gh
G= gpe / mh
H=gpe / mg

7 0
3 years ago
Read 2 more answers
What is the sidereal period used in Kepler’s third law?
klasskru [66]
His law exaplins/shows that the average distance of a planet from the Sun cubed is directly proportional to the orbital period squared.
3 0
3 years ago
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